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4 years ago

Which number can you add to get a sum of 0 A . -10 B. -5 C. 0 D. 5

1 answer:

4 0

To figure this one out, my pre calculus teacher said to "plug and chug". Since you are adding each number to itself, A=-20, B=-10, C=0, and D=10. So your answer would be C.0

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Help, write an expression in the simplest form that represents the area (in square feet) of the banner.

k0ka [10]

Answer:

3 (3 + x) is equal to the banner

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3 years ago

Read 2 more answers

There are 25 classes participating in Field Day. Each activity lasts 10 minutes. if field day starts at 9 am what time will all

melamori03 [73]

................................ the answer is 1500

7 0

3 years ago

Solve the triangles below

nataly862011 [7]

I guess you want the missing angle.
The missing angle is 83

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3 years ago

In this assignment, you will explore how the volume is changed when you adjust the height or radius of a cylinder. You will drag

lesantik [10]

The volume of a cylinder changes when you adjust the height or radius as the volume either increases or reduces.

<h3>How to illustrate the volume?</h3>

Let's assume that the height and radius are 14cm and 7cm. The volume will be:

= πr²h

= 3.14 × 7² × 14

= 2154.04cm³

When the radius and height are reduced to 5cm and 9cm, the volume will be:

= πr²h

= 3.14 × 5² × 9

= 706.5cm³

This illustrates that the volume reduces when the height and radius reduces.

Learn more about volume on:

brainly.com/question/1972490

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8 0

2 years ago

A computer programmer has an 18% chance of finding a bug in any given program. What is the probability that she does not find a

LiRa [457]

Answer: $(\frac{41}{50})^{10}$

Step-by-step explanation:

Since, According to the question,

The chance of finding a bug in a program = 18%

Thus, the probability of finding the bug in first attempt = $\frac{18}{100}=\frac{9}{50}$

⇒ The probability of not finding any bug in first attempt = $1 - \frac{9}{50} =\frac{41}{50}$

Similarly, in second attempt , third attempt, fourth attempt_ _ _ _ _ tenth attempt, the probability of not finding any bug is also equal to $\frac{41}{50}$

Thus, the probability that she does not find a bug within the first 10 programs she examines

= $\frac{41}{50}\times \frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}\times\frac{41}{50}$

= $(\frac{41}{50})^{10}$

5 0

3 years ago